# Rutherford alpha particle scattering

• Tyst
In summary, Dick made up equations to solve for the scattering at different angles. He found that at 20 degrees, there are 200 scattered alpha particles.

## Homework Statement

A parallel beam of alpha particles with fixed kinetic energy is normally incident on a piece of gold foil.
a) if 100 alpha particles per minute are detected at 20 degrees, how many will be counted at 40, 60, 80 and 100 degrees?
b) If the kinetic energy of the incident alpha partilces is doubled, how many scattered alpha particles will be observed at 20 degrees.
The density of the gold foil is given to be 19.3 g/cm^3

2. Homework Equations / Attempt at solution

N = k/[sin(theta/2)^4] ... My question is in relation to the relevant equations. I attempted to use this equation (substituting values given for scattering at 20 degrees to find 'k') for part a), though i am fairly sure there is more to it than this, could someone please point me in the right direction with regard to other equations relevant to this problem? I'm afraid i am lacking a text and have had no luck with google!

a.) 40 - 50
60 - 25
80 - 10
100 - 5

b.) 2x 200 2He4

Thank you for the response .ultimate, i was wondering if you could tell me how you came to get those answers? Perhaps you could tell me what equations were used, or how you came to that conclusion?

Thank you

Tyst said:
Thank you for the response .ultimate, i was wondering if you could tell me how you came to get those answers? Perhaps you could tell me what equations were used, or how you came to that conclusion?

Thank you

I don't think he used 'equations'. He made them up. It sounds like you are doing exactly the right thing for part a). Do b) in more or less the same way - but now you'll need to know the dependence of k on the kinetic energy. Here's a reference:

http://hyperphysics.phy-astr.gsu.edu/hbase/rutsca.html

Thanks Dick